\(I_{1}\;+\;I_{2}\;-\;0.005\; =\; 0\)
\(\frac{6\;-\;V_{x}}{1000}\;+\;\frac{7.2\;-\;V_{x}}{1000}\;-\;0.005\;=\;0\)
\({6\;-\;V_{x}}\;+\;{7.2\;-\;V_{x}}\;-\;5\;=\;0\)
\({-2V_{X}\;+\;(7.2+6)\;-\;5\;=\;0\)
\({-2V_{X}\;+\;8.2\;=\;0\)
\({2V_{X}\;=\;8.2\)
\(V_{x}\;=\frac{8.2}{2}\)
\(V_{x}\;=\;4.1\)
Plugging the value for Vx back into the expression for the I1:
\(I_{1}\;=\;\frac{6\;-\;4.1}{1000}\)
\(I_{1}\;=\;\frac{1.9}{1000}\)
\(I_{1}\;=\;{1.9}\; milliamps\)
You have to apply KVL consistently as you make a complete circuit through the network. If you cross a voltage source from negative to positive, you add that voltage otherwise you subtract it. If you cross a passive component (resistor, in this case) in the direction of the current you are assuming, you subtract the voltage otherwise you add it. If you move from a voltage as Node 1 to a voltage at Node 2, you subtract (V1 - V2).great, thank you hgmjr!
I tried solve the problem with KVL, what's wrong with my solution below?
let "I" be the current from the 6v battery and "Vx" be the voltage drop at the transistor.
1000I + Vx + 6 = 0 (1)
Thank you so much, that's very helpful!......
You can chose which direction you go around a given loop and you can choose whether you are going to add voltage drops or voltage gains as you go around it. But whatever you choose, you must then apply it consistently. You can make different choices on the next loop, but on each loop you must apply the choices made for that loop consistently.
Also, do yourself a really, really big favor. Start using units consistently and religiously throughout your work from beginning to end on every line. Tracking units is perhaps the single most valuable error detecting tool you have, because most (not all, but most) errors you make will screw up the units. But you can only detect those errors if the units are there to get screwed up.
\(V_{x}\;=\;\frac{\frac{6}{1000}\;+\;\frac{7.2}{1000}\;-\;0.005}{\frac{1}{1000}\;+\;\frac{1}{1000}}\)
\(\;\)
Don't spend time memorizing a bunch of different forumulas for this and that. Focus on understanding the fundamentals and being able to apply them with skill and confidence. Then those special case formulas that you end up constructing repeatedly you will naturally start shortcutting and using.That's a nice theorem, looks very handy.
Almost all scientists and engineers use conventional current exclusively. In nearly all cases it is equivalent and makes no physical difference, but it makes the math a lot cleaner and more consistent. There are exceptions and people that work in those fields commonly think in terms of whichever flow is important for what they do; for some it is electron flow because they are dealing with electron beams, for others it is positive current flow because they are dealing with beams of positively charged ions, others, such as those that work with semiconductors at the lowest levels, learn to think in terms of "majority" and "minority" carriers which could be electrons and "holes" in one region and the reverse, holes and electrons, in another. But unless you end up in one of these specialties, you will be far better served by sticking with conventional current flow.However, I still have a question regarding the current analysis. When I am doing mesh current analysis, I found it's much easier to use conventional current flow rather than the flow of electrons, even though I don't have any prior electrical knowledge. I figured that is probably because the signs of batteries make the flow of electrons very unintuitive. However, is there any trick in using the flow of electrons for current analysis? (I kind of wanna learn this way because it's more scientific)
That's great a suggestion.Don't spend time memorizing a bunch of different forumulas for this and that. Focus on understanding the fundamentals and being able to apply them with skill and confidence. Then those special case formulas that you end up constructing repeatedly you will naturally start shortcutting and using.
I kind of get it after spending many hours with the circuit from the worksheet. I figured that when using the flow of electrons, the current from negative to positive is a drop and from positive to negative is a charge as opposed to the other way around in the conventional current. Thus, in the flow of electrons, the negative side of a battery (if there's only one power supply) represents the highest potential and the positive side of the battery represent the lowest potential. Correct me if I am wrong.It's all about your reference point and thinking habits. Pick up a simple schematic and redraw it with negative as the "supply" and go through the process of analysing the circuit under those conditions. After a few of these, you will be able to do it without redrawing the circuit.
As a generalization (and almost certainly too strong a one), texts written from a "technician" perspective tend to me more mixed with many using electron flow and some using conventional flow. I have never seen an "engineering" text that used electron flow (but I'm sure that there have to be some). I think the difference stems from the level of math and theory that each type delves into. In the first type, formulas are often presented with hand-wavy explanations for where they come from and it is easier to visualize the explanations if you picture electrons moving about. In the second type, formulas are the product of pretty rigorous mathematical development from fundamental principles. In fact, it is the nature of the math and not the intended audience that determines which group I would place it in and "technician" and "engineer" are simply convenient qualitative labels. As such, it is very important to use a system that is very mathematically self-consistent, and that means conventional flow.That's great a suggestion.
I think it'd be better for me to able to use the two ways of current flow interchangeably, since different books use different current flows.
Another reason that why I want to get familiar with electron flow is because the online book here is actually using the flow electron.
If you want to use electron flow, then you have to keep a bunch of rules in mind. Yes, for electrons, the negative terminal of the battery is the higher potential (note that this is true for conventional flow if the charge involved is negatively charged) and the electrons move from the higher (negative) potential to the lower (positive) potential. But you are still stuck with the voltages as assigned for conventional flow. So a drop in potential is a gain in voltage. Thus when electron current flows through a resistor, it gains voltage (because losing potential equates to a higher voltage). That's part of what makes working with electron current problematic because you have to apply rules that are not self-consistent. Now, as with anything, once you use it enough it becomes second-nature and the inconsistencies seem to fade away and disappear because you are able to simply deal with them unconsciously.I kind of get it after spending many hours with the circuit from the worksheet. I figured that when using the flow of electrons, the current from negative to positive is a drop and from positive to negative is a charge as opposed to the other way around in the conventional current. Thus, in the flow of electrons, the negative side of a battery (if there's only one power supply) represents the highest potential and the positive side of the battery represent the lowest potential. Correct me if I am wrong.
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